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I need help understanding this corollary of the Poincaré-Bendixson Theorem from Differential Equations, Dynamical Systems, and an Introduction to Chaos, by Hirsch, Smale and Devaney.

Let $\omega(X),\alpha(X)$ be the set of limit points of a solution as $t \to \infty$ and $t \to -\infty$ respectively. Recall that a limit cycle is a closed orbit $\gamma$ such that $\gamma \subset \omega(X)$ or $\gamma \subset \alpha(X)$ for some $X \notin \gamma$.

Suppose that $K$ is positively invariant, closed and bounded. If $X \in K$ then $\omega(X)$ must also lie in $K$. Thus $K$ must contain either an equilibrium point or a limit cycle.

Corollary 2. A closed and bounded set $K$ that is positively or negatively invariant contains either a limit cycle or an equilibrium point.

What I don't understand is that shortly before stating that corollary, the book remarks that:

Not all closed orbits have this property [of being a limit cycle]. For example, in the case of a linear system with a center at the origin in $\mathbb{R}^2$, the closed orbits that surround the origin have no solutions approaching them and so are not limit cycles.

So wouldn't each one of those orbits constitute closed and bounded sets that are positively and negatively invariant and contain neither a limit cycle or an equilibrium?

As always, I'd appreciate any help in understanding this.

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  • $\begingroup$ A center is an equilibrium point..... every linear system has an equilibrium point at the origin.... $\endgroup$ – symplectomorphic Dec 17 '17 at 6:50
  • $\begingroup$ @symplectomorphic I don't understand, how does the origin figure into it? I'm saying the circles themselves are each closed bounded sets that contain neither a limit cycle nor an equilibrium point (although they do enclose an area that has an equilibrium point, of course). $\endgroup$ – mlaci Dec 17 '17 at 6:56
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    $\begingroup$ I thought you meant the disks, not the orbits themselves. The statement of the corollary is indeed wrong as stated: the authors are tacitly assuming that the topological boundary of $K$ doesn't contain a closed orbit (look at the picture above the corollary). If you take $K$ to be a circle in the center example, this condition is violated. For a more precise statement, see theorem 1.179 here: www4.ncsu.edu/~schecter/ma_732_sp13/p-b.pdf $\endgroup$ – symplectomorphic Dec 17 '17 at 7:41
  • $\begingroup$ @symplectomorphic The picture you mentioned is related to an example given on the previous page of the book, but I suppose you may be right that it was implied that the set didn't contain a closed orbit, although that just seems a bit odd. Thank you. If you'd like to post your comment as answer, I'll mark it as accepted. $\endgroup$ – mlaci Dec 17 '17 at 17:52

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